In order to solve this problem for a specific function of dφ/dt (time rate of latitude change), you must specify the latitude (or declination) of the sun. If the latitude is zero (sun over the equator), then the task is impossible. If the sun isn't on the equator, then there is some rate of dφ/dt that will keep you on the slanted terminator for a while. If the sun is south of the equator, then you'd go north at sunrise or south at sunset. If the sun is north of the equator, then you'd go south at sunrise or north at sunset.
t = observer's sidereal time
α = sun's right ascension
δ = sun's declination
φ = observer's latitude
E = sun's elevation
E = ArcSin{ sin(φ) sin(δ) + cos(φ) cos(δ) cos(t−α) } = 0
φ = Arctan{ −cos(t−α) / tan(δ) }
Finding dφ/dt…
u = −cos(t−α) / tan(δ)
du/dt = sin(t−α) / tan(δ)
dφ/dt = d/dt Arctan u = (du/dt) / (1+u²)
dφ/dt = sin(t−α) / [ tan(δ) + cos²(t−α) / tan(δ) ]
Remember that you must convert t from time rate to angle rate. The conversion is 15 degrees per hour. To get the speed relative to Earth's surface,
v = k R (dφ/dt)
where R is Earth's radius and φ is the observer's latitude in radians. The (k) is a conversion factor that I put in to account for any necessary changing of units, such as changing (t) from angle to time and changing φ from degrees to radians.
This answer will be approximate, since α and δ (the right ascension and declination of the sun) are also functions of time.
Example.
Let's suppose that δ=23.5° and φ=0 at time t=0. It's sunset, and the sun is north of the equator, so to keep it there the observer must travel north at some speed or other. Before we attempt to calculate that speed, however, we must determine the hour angle of the sun, relative to the observer, at t=t₀=0.
E = ArcSin{ sin(0) sin(23.5°) + cos(0) cos(23.5°) cos(t₀−α) } = 0
t₀−α = ±90°
Since we're beginning on the equator at sunset, t₀−α = +90°.
So the rate of change of latitude at time t=0 is
dφ/dt = sin(90°) / [ tan(23.5°) + cos²(90°) / tan(23.5°) ]
dφ/dt = 1 / [ tan(23.5°) + 0 ] ≈ 2.2998
The units of dφ/dt are degrees of latitude per degree of time-converted-to-angle. We would multiply this figure by 15 to get dφ/dt in degrees of latitude per hour of time.
dφ/dt = 34.3976 degrees per hour = 0.602097 radians per hour
R = 6378000 meters
v(at time t=0) = R (dφ/dt) = 3840000 meters per hour
v₀ = 1067 m/s
That speed is not constant. It changes with latitude and, hence, with time. The function v(t) is transcendental, so you'll have to find it by numerical integration, stepping over time and solving, like this:
t₁ = t₀ + Δt
φ₁ = φ₀ + Δt sin(t₀−α) / [ tan(δ) + cos²(t₀−α) / tan(δ) ]
v₁ = π R (φ₁−φ₀) / 43200
t₂ = t₁ + Δt
φ₂ = φ₁ + Δt sin(t₁−α) / [ tan(δ) + cos²(t₁−α) / tan(δ) ]
v₂ = π R (φ₂−φ₁) / 43200
And so on. The values for t in the trigonometric functions are in angular units, but otherwise they are in time units. Convert as necessary. Δt should be chosen as small as is practical. The values for v will be in meters per second.
Initial conditions: φ₀ = 0, δ = +23.5°, t₀−α = +90°
time in sec, latitude in degrees, speed in m/s
0, 0, 1066.7
1000, 9.486, 1035.0
2000, 18.434, 949.9
3000, 26.462, 834.6
4000, 33.407, 712.1
5000, 39.279, 597.3
6000, 44.183, 497.1
7000, 48.258, 412.8
8000, 51.643, 343.2
9000, 54.460, 285.9
10000, 56.809, 238.8
11000, 58.773, 199.7
12000, 60.416, 167.1
13000, 61.790, 139.5
14000, 62.934, 115.9
15000, 63.882, 95.43
16000, 64.656, 77.39
17000, 65.277, 61.24
18000, 65.760, 46.52
19000, 66.116, 32.85
20000, 66.353, 19.92
21000, 66.475, 7.410
21600, 66.5, 0
